Results 1 to 7 of 7

Thread: Prove identity

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Prove identity

    Prove that $\displaystyle \arctan x+\arctan y\equiv \arctan\frac{x+y}{1-xy}$.
    use this relationship to show that, if $\displaystyle \arctan x+\arctan y+arctan z=\frac{\pi}{2}$ then $\displaystyle xy+yz+xz=1$
    I have done the first part:
    Let $\displaystyle \arctan x=\theta$ and $\displaystyle \arctan y=\phi$
    then for LHS
    $\displaystyle \equiv \theta +\phi$
    $\displaystyle \equiv \arctan(\tan(\theta+\phi))$
    $\displaystyle \equiv \arctan(\frac{tan \theta+\tan \phi}{1-\tan \theta \tan \phi})$
    $\displaystyle \equiv \arctan(\frac{x+y}{1-xy})\equiv RHS$
    now my problem is the second part, after i subsitute $\displaystyle \arctan(\frac{x+y}{1-xy})$ into $\displaystyle \arctan x+\arctan y+arctan z=\frac{\pi}{2}$ to get $\displaystyle \arctan(\frac{x+y}{1-xy})+\arctan z=\frac{\pi}{2}$, i don't know what to do.
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    Let $\displaystyle \frac{x+y}{1-xy}=w$

    Apply again the first identity for $\displaystyle \arctan w+\arctan z$ and replace back w.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    thanks! i did that, and let $\displaystyle arctan w=\omega$ and $\displaystyle arctan z=\gamma$
    so i had $\displaystyle \omega+\gamma=\frac{\pi}{2}$
    $\displaystyle arctan(tan(\omega+\gamma)=\frac{\pi}{2}$
    $\displaystyle arctan(\frac{w+z}{1-wz})=\frac{\pi}{2}$
    $\displaystyle arctan(\frac{\frac{x+y}{1-xy}+z}{1-z(\frac{x+y}{1-xy})})=\frac{\pi}{2}$
    $\displaystyle arctan(\frac{x+y+z-xyz}{1-xy-xz-yz})=\frac{\pi}{2}$
    now i'm stuck, $\displaystyle tan\frac{\pi}{2}$ doesn't give a value
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    If $\displaystyle \arctan\frac{a}{b}=\frac{\pi}{2}$ then b must be 0.

    In this case $\displaystyle 1-xy-xz-yz=0\Rightarrow xy+xz+yz=1$

    Edited.
    Last edited by red_dog; Jul 30th 2009 at 12:45 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    i thinking something along that line but didn't have any confirmation
    thanks!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Quote Originally Posted by red_dog View Post
    If $\displaystyle \arctan\frac{a}{b}=0$ then b must be 0.

    In this case $\displaystyle 1-xy-xz-yz=0\Rightarrow xy+xz+yz=1$
    Whoops! I think you mean 'If $\displaystyle \arctan\frac{a}{b}=\frac{\pi}{2}$ then b must be 0'.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    Quote Originally Posted by Grandad View Post
    Whoops! I think you mean 'If $\displaystyle \arctan\frac{a}{b}=\frac{\pi}{2}$ then b must be 0'.

    Grandad
    Sorry! You're right! I've edited.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. prove identity help!
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: Dec 2nd 2010, 10:00 AM
  2. prove the identity
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jun 7th 2010, 05:03 PM
  3. Prove this identity?
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: Apr 5th 2010, 05:29 PM
  4. Prove that each of the following is an identity
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Apr 18th 2009, 02:39 PM
  5. Prove the Identity
    Posted in the Trigonometry Forum
    Replies: 10
    Last Post: Feb 25th 2008, 05:29 PM

Search Tags


/mathhelpforum @mathhelpforum